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College Physics by Openstax Chapter 6 Problem 27

The radius and centripetal acceleration of a bobsled turn on an ideally banked curve


Problem:

(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?


Solution:

Part A

For ideally banked curved, the ideal banking angle is given by the formula tanθ=v2rg\displaystyle \tan \theta = \frac{v^2}{rg}. We can solve for rr in terms of all the other variables, and we should come up with

r=v2gtanθr = \frac{v^2}{g \tan \theta}

We are given the following values:

  • ideal banking angle, θ=75.0 \displaystyle \theta = 75.0\ ^\circ
  • linear speed, v=30.0 m/s\displaystyle v=30.0\ \text{m/s}
  • acceleration due to gravity, g=9.81 m/s2\displaystyle g=9.81\ \text{m/s}^2

If we substitute all the given values into our formula for rr, we have

r=v2gtanθr=(30.0 m/s)2(9.81 m/s2)(tan75)r=24.5825 mr=24.6 m  (Answer)\begin{align*} r & = \frac{v^2}{g \tan \theta} \\ \\ r & = \frac{\left( 30.0\ \text{m/s} \right)^2}{\left( 9.81\ \text{m/s}^2 \right)\left( \tan 75^\circ \right)} \\ \\ r & = 24.5825\ \text{m} \\ \\ r & = 24.6\ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The radius of the ideally banked curve is approximately 24.6 m24.6\ \text{m}.

Part B

The centripetal acceleration aca_c can be solved using the formula

ac=v2ra_c = \frac{v^2}{r}

Substituting the given values, we have

ac=v2rac=(30.0 m/s)224.5825 mac=36.6114 m/s2ac=36.6 m/s2  (Answer)\begin{align*} a_c & = \frac{v^2}{r} \\ \\ a_c & = \frac{\left( 30.0\ \text{m/s} \right)^2}{24.5825\ \text{m}} \\ \\ a_c & = 36.6114\ \text{m/s}^2 \\ \\ a_c & = 36.6 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal acceleration is about 36.6 m/s236.6\ \text{m/s}^2.

Part C

To know how large is the computed centripetal acceleration, we can compare it with that of acceleration due to gravity.

acg=36.6114 m/s29.81 m/s2=3.73\frac{a_c}{g} = \frac{36.6114\ \text{m/s}^2}{9.81\ \text{m/s}^2} = 3.73

The computed centripetal acceleration is 3.73 times the acceleration due to gravity. That is ac=3.73ga_c = 3.73g.

This does not seem too large, but it is clear that bobsledders feel a lot of force on
them going through sharply banked turns!


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College Physics by Openstax Chapter 6 Problem 26

The Ideal Speed on a Banked Curve


Problem:

What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?


Solution:

The formula for the ideal speed on a banked curve can be derived from the formula of the ideal angle. That is, starting from tanθ=v2rg\tan \theta = \frac{v^2}{rg}, we can solve for vv.

v=rgtanθv = \sqrt{rg \tan \theta}

For this problem, we are given the following values:

  • radius of curvature, r=100 mr=100\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g=9.81\ \text{m/s}^2
  • banking angle, θ=20.0\theta = 20.0 ^\circ

If we substitute the given values into our formula, we have

v=rgtanθv=(100 m)(9.81 m/s2)(tan20.0)v=18.8959 m/sv=18.9 m/s  (Answer)\begin{align*} v = & \sqrt{rg \tan \theta} \\ \\ v = & \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right) \left( \tan 20.0 ^\circ \right) } \\ \\ v = & 18.8959\ \text{m/s} \\ \\ v = & 18.9\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ideal speed for the given banked curve is about 18.9 m/s18.9\ \text{m/s}.


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College Physics by Openstax Chapter 6 Problem 25

The ideal banking angle of a curve on a highway


Problem:

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?


Solution:

The ideal banking angle (meaning there is no involved friction) of a car on a curve is given by the formula:

θ=tan1(v2rg)\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)

We are given the following values:

  • radius of curvature, r=1.20 km×1000 m1 km=1200 m\displaystyle r = 1.20\ \text{km} \times \frac{1000\ \text{m}}{1\ \text{km}} = 1200\ \text{m}
  • linear velocity, v=105 km/h×1000 m1 km×1 h3600 s=29.1667 m/s\displaystyle v=105\ \text{km/h}\times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} = 29.1667\ \text{m/s}
  • acceleration due to gravity, g=9.81 m/s2\displaystyle g = 9.81\ \text{m/s}^2

If we substitute these values into our formula, we come up with

θ=tan1(v2rg)θ=tan1[(29.1667 m/s)2(1200 m)(9.81 m/s2)]θ=4.1333θ=4.13  (Answer)\begin{align*} \theta & = \tan^{-1} \left( \frac{v^2}{rg} \right) \\ \\ \theta & = \tan^{-1} \left[ \frac{\left( 29.1667\ \text{m/s} \right)^2}{\left( 1200\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\ \theta & = 4.1333 ^\circ \\ \\ \theta & = 4.13 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ideal banking angle for the given highway is about 4.13 4.13 ^\circ.


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College Physics by Openstax Chapter 6 Problem 24

Centripetal Force of a Rotating Wind Turbine Blade


Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.


Solution:

We are given the following values:

  • radius, r=100 mr=100\ \text{m}
  • angular velocity, ω=0.5 rev/sec×2π rad1 rev=3.1416 rad/sec\omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}
  • mass, m=4 kgm=4\ \text{kg}

Centripetal force FcF_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity vv and has magnitude Fc=macF_c = m a_c which can also be expressed as

Fc=mv2ror Fc=mrω2F_c = m \frac{v^2}{r} \quad \text{or} \quad \ F_c = mr \omega^2

For this particular problem, we are going to use the formula Fc=mrω2F_c = mr \omega^2. If we substitute the given values, we have

Fc=mrω2Fc=(4 kg)(100 m)(3.1416 rad/sec)2Fc=3947.8602 NFc=4×103 N  (Answer)\begin{align*} F_c & =mr \omega^2 \\ \\ F_c & = \left( 4\ \text{kg} \right)\left( 100\ \text{m} \right)\left( 3.1416\ \text{rad/sec} \right)^2 \\ \\ F_c & = 3947.8602\ \text{N} \\ \\ F_c & = 4 \times 10^3\ \text{N}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal force on the end of the wind turbine blade is approximately 4×103 N4 \times 10^3\ \text{N}.


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College Physics by Openstax Chapter 6 Problem 23

The centripetal force of a child riding a merry-go-round


Problem:

(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.


Solution:

Part A

We are given the following values: m=22.0 kgm=22.0\ \text{kg}, ω=40.0 rev/min\omega = 40.0\ \text{rev/min}, and r=1.25 mr=1.25\ \text{m}. We are asked to solve for the centripetal force, FcF_c.

Centripetal force FcF_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity vv and has magnitude Fc=macF_c=m a_c, which can also be expressed as Fc=mv2rF_c = m \frac{v^2}{r} or Fc=mrω2F_c = m r \omega ^2. Basing from the given values, we are going to solve the problem using the formula

Fc=mrω2F_c = m r \omega ^2

First, we need to convert the angular velocity ω\omega to rad/sec\text{rad/sec} for unit homogeneity.

40 rev/min×2π rad1 rev×1 min60 sec=4.1888 rad/sec40\ \text{rev/min} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} = 4.1888\ \text{rad/sec}

Now, we can substitute the given values into our formula.

Fc=mrω2Fc=(22.0 kg)(1.25 m)(4.1888 rad/sec)2Fc=482.5162 NFc=483 N  (Answer)\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 1.25\ \text{m}\right) \left( 4.1888\ \text{rad/sec}\right)^2 \\ \\ F_c & = 482.5162\ \text{N} \\ \\ F_c & = 483\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let us convert the angular velocity to radians per second.

3.00 rev/min×2π rad1 rev×1 min60 sec=0.3142 rad/sec3.00\ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}}=0.3142 \ \text{rad/sec}

Now, we can substitute the given values

Fc=mrω2Fc=(22.0 kg)(8.00 m)(0.3142 rad/sec)2Fc=17.3750 NFc=17.4 N  (Answer)\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 8.00\ \text{m}\right) \left( 0.3142\ \text{rad/sec}\right)^2 \\ \\ F_c & = 17.3750\ \text{N} \\ \\ F_c & = 17.4\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For the first centripetal force we solved in Part A,

Fcw=483 N(22 kg)(9.81 m/s2)=2.24\frac{F_c}{w} = \frac{483\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 2.24

The centripetal force is 2.24 times the weight of the child.

For the centripetal force we solved in Part B, we have

Fcw=17.4 N(22 kg)(9.81 m/s2)=0.0806\frac{F_c}{w} = \frac{17.4\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 0.0806

The centripetal force is only about 8% of the child’s weight.


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College Physics by Openstax Chapter 6 Problem 21

Centripetal acceleration of an amusement park ride shaped like a Viking ship


Problem:

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. The speed at the bottom of the arc is 23.4 m/s.

(a) What is the centripetal acceleration at the bottom of the arc?

(b) Draw a free-body diagram of the forces acting on a rider at the bottom of the arc.

(c) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(d) Discuss whether the answer seems reasonable.


Solution:

Problem 6-20: The centripetal acceleration of the commercial jet’s tires, and the force of a determined bacterium in it


At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.

(a) At how many rev/min are the tires rotating?

(b) What is the centripetal acceleration at the edge of the tire?

(c) With what force must a determined 1.00×10−15 kg bacterium cling to the rim?

(d) Take the ratio of this force to the bacterium’s weight.


Solution:

We are given the following quantities: linear speed, v=60.0 m/s v=60.0 \ \text{m/s}, radius is half the diameter, r=0.425 m r=0.425 \ \text{m}.

Part A

We can compute the angular velocity based on the given using the formula, ω=vr \displaystyle \omega = \frac{v}{r}.

ω=vrω=60.0 m/s0.425 mω=141.1765 rad/sec\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{60.0 \ \text{m/s}}{0.425 \ \text{m}} \\ \\ \omega & = 141.1765 \ \text{rad/sec} \end{align*}

Now, we can convert this into the required unit of rev/min.

ω=141.1765 radsec×1 rev2π rad×60 sec1 minω=1348.1363 rev/minω=1.35×103 rev/min  (Answer)\begin{align*} \omega & = 141.1765\ \frac{\text{rad}}{\text{sec}} \times \frac{1\ \text{rev}}{2\pi\ \text{rad}} \times \frac{60\ \text{sec}}{1\ \text{min}} \\ \\ \omega & = 1348.1363 \ \text{rev/min} \\ \\ \omega & = 1.35 \times 10^{3} \ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The centripetal acceleration at the edge of the tire can be computed using the formula, ac=rω2 a_{c} = r \omega ^{2}.

ac=rω2ac=(0.425 m)(141.1765 rad/sec)2ac=8470.5918 m/s2ac=8.47×103 m/s2  (Answer)\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.425\ \text{m} \right) \left(141.1765\ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 8470.5918 \ \text{m/s}^2 \\ \\ a_{c} & = 8.47 \times 10 ^{3} \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

From the second law of motion, the force is equal to the product of the mass and the acceleration. In this case, we are going to use the formula, Fc=mac F_c = m a_c . We are given the mass to be m=1.00×1015 kgm=1.00 \times 10 ^{-15}\ \text{kg} , and the centripetal acceleration is solved in Part B.

Fc=macFc=(1×1015 kg)(8470.5918 m/s2)Fc=8.4705918×1012 kg m/s2Fc=8.47×1012 N  (Answer)\begin{align*} F_c & = ma_c \\ \\ F_c & = \left( 1 \times 10^{-15}\ \text{kg}\right) \left(8470.5918 \ \text{m/s}^2\right) \\ \\ F_c & = 8.4705918 \times 10 ^{-12}\ \text{kg m/s}^2 \\ \\ F_c & = 8.47 \times 10^{-12} \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The ratio of this force, Fc F_c to the weight of the bacterium is

Fcmg=8.4705819×1012 N(1×1015kg)(9.81 m/s2)Fcmg=863.4640Fcmg=863  (Answer)\begin{align*} \frac{F_c}{mg} & = \frac{8.4705819 \times 10 ^{-12}\ \text{N}}{\left( 1 \times 10^{-15} \text{kg} \right)\left(9.81 \ \text{m/s}^2 \right)} \\ \\ \frac{F_c}{mg} & = 863.4640 \\ \\ \frac{F_c}{mg} & = 863 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-19: The angular velocity of an “artificial gravity”


A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?


Solution:

We are given the following quantities:

radius=diameter2=200 m2=100 m\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
centripetal acceleration,ac=9.80 m/s2\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

ac=rω2a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

ω=acr\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

ω=acrω=9.80 m/s2100 mω=0.313 rad/sec  (Answer)\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\ \omega & = 0.313 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-18: The linear speed of an ultracentrifuge and Earth in its orbit


Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).


Solution:

Part A

We are given a linear speed of an ultracentrifuge of 0.50 km/s0.50\ \text{km/s}. We are asked to verify this value if we are given a radius of r=0.100 mr=0.100\ \text{m} and angular velocity of ω=50000 rev/min \omega = 50000 \ \text{rev/min}. We are going to use the formula

v=rωv = r \omega

Since we are given a linear speed in km/s\text{km/s}, we are going to convert the radius to km\text{km}, and the angular velocity to rad/sec\text{rad/sec}

r=0.100 m×1 km1000 m=0.0001 kmr=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km}
ω=50000 rev/min×2π rad1 rev×1 min60 sec=5235.9878 rad/sec\omega = 50000 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} =5235.9878\ \text{rad/sec}

Now, we can substitute these into the formula

v=rωv=(0.0001 km)(5235.9878 rad/sec)v=0.5236 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 0.0001 \ \text{km} \right)\left( 5235.9878 \ \text{rad/sec} \right) \\ \\ v & = 0.5236 \ \text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This value is about 0.500 km/s.

Part B

From Table 6.2 of the book

ParentSatelliteAverage orbital radius r(km)Period T(y)r3 / T2 (km3 / y2)
SunEarth1.496×1081.496 \times 10^{8} 13.35×10243.35 \times 10^{24}

Using the same formulas we used in Part A, we can solve for the linear velocity of the Earth around the sun. The radius is

r=1.496×108 kmr=1.496 \times10^{8} \ \text{km}

The angular velocity is

ω=1 revyear×2π rad1 rev×1 year365.25 days×1 day24 hours×1 hour3600 secω=1.9910×107 rad/sec\begin{align*} \omega & = 1 \ \frac{\text{rev}}{\text{year}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{year}}{365.25 \ \text{days}} \times \frac{1\ \text{day}}{24\ \text{hours}}\times \frac{1\ \text{hour}}{3600\ \text{sec}} \\ \\ \omega & = 1.9910 \times 10^{-7}\ \text{rad/sec} \end{align*}

The linear velocity is

v=rωv=(1.496×108 km)(1.9910×107) rad/secv=29.7854 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 1.496\times 10^{8}\ \text{km} \right)\left( 1.9910 \times 10 ^ {-7} \right) \ \text{rad/sec}\\ \\ v & = 29.7854\ \text{km/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The linear velocity is about 30 km/s.


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Problem 6-17: The acceleration due to gravity at the position of a satellite located above the Earth


What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?


Solution:

The acceleration due to gravity of a body and the Earth is given by the formula

g=GMr2g= G \frac{M}{r^2}

where GG is the gravitational constant, MM is the mass of the Earth, and rr is the distance of the object to the center of the Earth. We know that the approximate radius of the Earth is r=6.3781×106 mr=6.3781 \times 10^6 \ \text{m} .

The percentage of the acceleration at 300 km above the Earth of the acceleration due to gravity at Earth’s surface is

(GMr2)2(GMr2)1×100%\displaystyle \frac{\left( \frac{GM}{r^2} \right)_2}{\left( \frac{GM}{r^2} \right)_1} \times 100\%

Note that the subscript 2 indicates the satellite located 300 km above the Earth, and the subscript 1 indicates the object at the Earth’s surface. Also, from the expression above, we can cancel GG and MM from the numerator and denominator because these are constants. So, we are down to

(1r2)2(1r2)1×100%=(r2)1(r2)2×100%\frac{\left( \frac{1}{r^2} \right)_2}{\left( \frac{1}{r^2} \right)_1} \times 100\% = \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\%

Substituting the values, we have

(r2)1(r2)2×100%=(6.3781×106 m)2(6.3781×106 m+300×103 m)2×100%=91.2172%=91.2%  (Answer)\begin{align*} \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\% & = \frac{\left( 6.3781 \times 10^6 \ \text{m} \right)^{2}}{\left( 6.3781 \times 10^6 \ \text{m}+300 \times 10^{3} \ \text{m} \right)^{2}} \times 100\% \\ \\ & = 91.2172\% \\ \\ & = 91.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The percentage of the acceleration at the Earth’s surface of the acceleration due to gravity at the position of a satellite located 300 km above the Earth is about 91.2%.


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